Question: Let $f(x)=2x^3+7x^2+4x-5$. On which intervals is $f$ decreasing? Choose 1 answer: Choose 1 answer: (Choice A) A $x>-\dfrac13$ only (Choice B) B $x>-2$ only (Choice C) C $x<-\dfrac13$ only (Choice D) D $-2<x<-\dfrac13$ only (Choice E) E The entire domain of $f$
Answer: We can analyze the intervals where $f$ is increasing/decreasing by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $f$ is $f'(x)=2(3x+1)(x+2)$. $f'(x)=0$ for $x=-2,-\dfrac13$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-2$ and $x=-\dfrac13$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $\begin{array}{rl} x<& \llap{-}2\end{array}$ $ \llap{-}2<x<-\frac{1}{3}$ $-\frac{1}{3}$ $ x>-\frac{1}{3}$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<-2$ $x=-3$ $f'(-3)=16>0$ $f$ is increasing $\nearrow$ $-2<x<-\dfrac13$ $x=-\dfrac23$ $f'\left(-\dfrac23\right)=-\dfrac83<0$ $f$ is decreasing $\searrow$ $x>-\dfrac13$ $x=0$ $f'(0)=4>0$ $f$ is increasing $\nearrow$ In conclusion, $f$ is decreasing over the interval $-2<x<-\dfrac13$ only.